# Tagged Questions

Modular arithmetic (clock arithmetic) is a system of integer arithmetic based on the congruence relation $a \equiv b \pmod{n}$ which means that $n$ divides $b-a$.

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### How many solutions does $x^2 = 1$ have in $\mathbb{Z}/m\mathbb{Z}$ in general?

As the title suggests, how many solutions does $x^2 = 1$ have in $\mathbb{Z}/m\mathbb{Z}$ in general?
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### Is it true that $\mathbb{Z}[i]/m\mathbb{Z}[i]$ has exactly $\text{N}(m)$ elements?

As the question title suggests, is it true that $\mathbb{Z}[i]/m\mathbb{Z}[i]$ has exactly $\text{N}(m)$ elements?
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### Are there infinitely many primes $n$ such that $\mathbb{Z}_n^*$ is generated by $\{ -1,2 \}$?

Let $n$ a prime, and let $\mathbb{Z}_n$ denote the integers modulo $n$. Let $\mathbb{Z}^*_n$ denote the multiplicative group of $\mathbb{Z}_n$ Are there infinitely many $n$ such that $\mathbb{Z}^*_n$ ...
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### Proving $133|a^{18}-b^{18}$ if $\gcd(a,133)=\gcd(b,133)=1$.

If $\gcd(a,133)=\gcd(b,133)=1$ then prove that $133|a^{18}-b^{18}$. Using Fermat theorem: $a^{132} \equiv 1\mod\ 133$ and $b^{132} \equiv 1\mod\ 133$, so $a^{132} \equiv b^{132}\mod\ 133$. What ...
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### Show that, of any set of $2^{n+1}-1$ positive integers numbers is possible choose $2^{n}$ elements such that their sum is divisible by $2^{n}$.

Show that, of any set of $2^{n+1}-1$ positive integers numbers is possible choose $2^{n}$ elements such that their sum is divisible by $2^{n}$. My approach: Let $a_1,a_2,a_3,\ldots,a_{2^{n+1}-1}$ ...
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### Why is $15m=0\pmod {18}$ equivalent to $3m=0 \pmod {18}$?

$m$ is some natural number. Why is $15m=0\pmod {18}$ equivalent to $3m=0 \pmod {18}$? This is my question basically. I don't understand why that's true.
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### Prove that $x \pmod n = 0$ if $x \pmod n = (k*x) \pmod n$ and $k \ne n$

I need a quick formal proof for : Prove that $x \pmod n = 0$ if $x \pmod n = (k*x) \pmod n$ and $k \ne n$ Thanks
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### $1989 \mid n^{n^{n^{n}}} - n^{n^{n}}$ for integer $n \ge 3$

Before anyone comments, yes this is kind of a duplicate of Prove that $1989\mid n^{n^{n^{n}}} - n^{n^{n}}$ . The problem that I'm having I don't see the $n=5$ as a counterexample. Also if anyone wants ...
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### Prove $\sum_{k=1}^{n} 2^n \text{ mod }k > 2n$ where $n > 1000$

This problem is taken from a Russian textbook of past Olympiads. Its statement looks like this : Given a natural number $n > 1000$ prove that $\sum_{k=1}^{n} 2^n \text{ mod }k > 2n$. ...
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### Why does a = b (mod n) iff a - b is divisible by n? [duplicate]

I am specifically asking why the statement $a \equiv b \;(\bmod\; n)$ is equivalent to the statement $a = b + kn$, where k is some positive integer. Why is it that the difference of a and b has to be ...
$3t_1 \equiv 1 \pmod 5$ $t_1 \equiv 2 \pmod 5$ how can we derive line 2 from line 1?